All,
I have a shell script which is attempting to both copy and install a .deb as part of a late_command. I used what I found from this link as a guide.
https://gist.github.com/moonwitch/11100762
Here are the echo commands appending things to the end of lubuntu.seed
echo "d-i preseed/late_command string \ " >> "$WORK_DIR/preseed/lubuntu.seed"
echo "cp /cdrom/pool/extras/my.deb . ; \ " >> "$WORK_DIR/preseed/lubuntu.seed"
echo "d-i preseed/late_command string in-target dpkg -i /cdrom/pool/extras/my.deb ; " >> "$WORK_DIR/preseed/lubuntu.seed"
What really honks me off is the cp command doesn't appear to leave the file on the target no matter what I do. There are some other things I would like to copy and unzip.
I'm working with Ubuntu 15 32-bit. Really don't care about tools that "used to work" with earlier releases. I have spent 3 days searching through every discussion and every example/tool was for a really old version of Ubuntu. Haven't found one which still works today. Some kind of run, but they don't fully function.
My current problem is due to the fact I'm not a packager person, but this has to get done and I didn't get out of the way fast enough.
What I need is an actual functioning cp statement which copies a file from the CD to the target and leaves it there through reboot. So far nothing has.
Thank you
This is the line I used for the late command:
d-i preseed/late_command string cp /cdrom/somefile /target/system/folder && cp /cdrom/someshellscript.sh /target/system/folder && chroot /target chmod a+x /target/system/folder/someshellscript.sh && chroot /target sh /system/folder/someshellscript.sh
I appended the line above to my preseed file. This line basically copies some script from the cdrom or iso to some folder in the target system. During installation the /target folder is the "target" systems root system (get it?).
So for me I copied all the files I need from the CD or iso to some folder like root or opt or even tmp in the target system. Then I changed root to /target and executed all the regular linux commands. Wrote a shell script that executes after all the copying is done to do what I want it to do.
Note:
Inside my shell script, I unset all of the installer's environment stuff:
unset DEBCONF_REDIR
unset DEBCONF_FRONTEND
unset DEBIAN_HAS_FRONTEND
unset DEBIAN_FRONTEND
Then proceed accordingly.
Hope this helps.
Related
When compiling, I always place the build in a separate directory. For example:
mkdir build
cd ./build
(cd ..; ./bootstrap)
../configure
make
Since I have plenty of RAM the aim is to compile on a TMPFS.
The script gets the name of the project, uses it for the name for the directory created in $XDG_RUNTIME_DIR/build and finally links it.
# setup-build.sh
#!/usr/bin/bash
set -e
my_project_name=$(basename $(pwd))
my_project_build_dir="$XDG_RUNTIME_DIR/build/$my_project_name"
mkdir -p $my_project_build_dir
ln -s "$my_project_build_dir" "$(pwd)/build"
The script runs without a problem. But, when I do cd ./build; ../configure it returns an error: bash: ../configure: No such file or directory. The file most certainly does exist, but Bash can't find it!
I altered the script to this:
#!/usr/bin/bash
set -e
my_project_src_dir="$(pwd)"
my_project_name="$(basename $(pwd))"
my_project_build_dir="$XDG_RUNTIME_DIR/build/$my_project_name"
mkdir -p "$my_project_build_dir"
ln -s "$my_project_build_dir" "$(pwd)/build"
cd "$my_project_build_dir"
echo "$my_project_src_dir" > "./project-src-dir.txt"
To compile I have to type cd ./build; $(cat ./project-src-dir.txt)/configure; make. This causes Bash complete to partial break, though. As in I can't TAB complete file names from $my_project_src_dir with this method, but TAB completion for arguments works fine. Ifautoconf is needed: (cd $(cat ./project-src-dir.txt); ./bootstrap). If anyone has any other ideas I would still prefer to be able to just do ../configure, although this will have to do for now.
Edit: Had to change my_project_name="$(basename '$my_project_src_dir') to my_project_name="$(basename $(pwd))" as it was taking '$my_project_src_dir' literally.
I am trying to compile grammar parser https://github.com/RichardMoot/Grail into Linux program according to instructions https://github.com/RichardMoot/Grail/blob/master/README and http://www.labri.fr/perso/moot/tutorial/install.html. There is manual how to create Linux executable from SWI-Prolog code http://www.swi-prolog.org/FAQ/UnixExe.html. All that is fine. But I can not find in the Makefile https://github.com/RichardMoot/Grail/blob/master/Makefile any compilation command. SWI-Prolo uses swipl command for compilation but this Makefile swipl calls only once - for displaying the version of the swipl.
I experience some hardship in installation and compilation, that is fine, I can execute/debug Makefile line by line and arrive at the result. But there is problem in my case - I can not see the ultimate goal in my makefile: which lines are responsible for the production of object files (if necessary) and which lines are responsible for the creation of the final Linux executable.
This is windowed program. The source code and documentation contains warnings about incompatibility with the SWI-Prolog 7, but that is fine, I can resolvem them myself, but as I said - I can not see the Makefile lines for creation of exe.
The source code is created by eminent scientist and I certainly don't want to disturb him by so low-level technical question. I would be happy if he continues work on theory and doesn't waste time on low level programming questions. Hope, that there are SWI-Prolog experts.
I am using latest (7.x) SWI-Prolog on Ubuntu 16.x and I have already installed all the mentioned prerequisites.
If you look closely at the provided Makefile, you'll find that the rules all and install are defined as follows (comments added by me):
all:
-cd source ; $(edit) g3 > g3.tmp # Replaces placeholders for your
# ... GRAIL_ROOT install directory.
-cd source ; mv -f g3.tmp g3 # Overwrites `g3` with the filled file.
cd source ; chmod a+x g3 # Makes it executable.
install: # Essentially copies all files to
-mkdir $(datarootdir) # ... your install directory.
-mkdir $(datadir)
cp -f $(images) $(datadir)
-mkdir $(bindir)
cp -f source/insertdot $(bindir)
chmod a+x $(bindir)/insertdot
cp -f $(resources) $(datadir)
cp -f source/*.pl $(bindir)
cp -f source/g3 $(bindir)
If you then do the common make && make install you'll end up with two folders installed in your Grail directory: bin and share. Inside the binary directory you'll have the g3 file that, regardless of being a SWI-Prolog source, has this initial line:
#!/usr/bin/swipl -q -g start -f
% [... prolog code.]
This header should allow your console terminal to determine what interpreter to use for this script (in this case, swipl). In my case, executing Grail with ./g3 returned a SWI-Prolog message indicating that wrong options/command arguments were used.
According to the man, Unix systems have to use option -s at the end of the header (but this didn't work either in my case):
From the manual:
-s file
Load file as a script. This option may be used from the shell to
make Prolog load a file before entering the toplevel.
It is also used to turn a file into an executable Prolog script
on Unix systems using the following first line
#!/usr/bin/swipl option ... -s
If you want to run this program, simply call the same command from your terminal:
swipl -q -g start -s g3
I'm new to shell programming and I'm trying to create a simple script that gives me some infos on the status of the machine (i.e date, time, users logged in etc) on Scientific Linux 6 (I know it's old, but the department of my university runs on it so there's no escaping)
Basically I've created my script "sysinfo.sh"
#!/bin/sh
....
exit 0
as root user I want to move it so that I can be able to execute it anywhere and I thought the right way to do it was
sudo mv sysinfo.sh usr/local/bin
but I get the error message
mv: cannot move `sysinfo.sh' to `usr/local/bin': No such file or directory
then I looked for the PATH and it gives me
$ echo $PATH
/u/geo2/sw//System/tools/bin:/usr/bin:/bin
What is the right place to move my script?
Best practice for these kind of manipulation or learning is to have scripts in your $HOME/bin directory.
mkdir $HOME/bin
export PATH=$PATH:$HOME/bin
mv sysinfo.sh $HOME/bin
chmod +x $HOME/bin/sysinfo.sh
If you anyway want to move it to /usr/local/bin, why not do that with:
sudo mv sysinfo.sh /usr/local/bin
chmod +x /usr/local/bin/sysinfo.sh
chmod command will make the script executable.
from chmod man:
x -- The execute/search bits.
The command that you posted indicates that you were trying to use the absolute path for copying, but you missed a leading slash --
the directory should be /usr instead of usr.
Try
sudo mv sysinfo.sh /usr/local/bin
Note that unless an absolute path is specified, the shell looks for the path relative to the current working directory.
In this case, the shell was looking for the subdirectory usr under the current directory which was not found;
hence the error message.
Thank you very much!
In the end, I didn't realize that the directory /usr/local/bin wasn't in the PATH
So i just needed to
export PATH=$PATH:/usr/local/bin
sudo mv sysinfo.sh /usr/local/bin
:D
I have dozens of scripts, all in different directories. (exported/expanded Talend jobs)
At this moment each job has 1 or 2 scripts, starting with the same lines, most important one:
CD ***path-to-script***
and several lines to set the Java path and start the job.
I want to create a script, which will be ran from all these scripts.
e.g.:
/scripts/talend.sh
And in all talend scripts, the first line will run /scripts/talend.sh, some examples of where these scripts are ran from:
/talend-job1_0.1/talend-job1_0.1/talend-job1/talend-job1.sh
/talend-task2_0.1/talend-task2_0.1/talend-task2/talend-task2.sh
/talend-job3_0.1/talend-job3_0.1/talend-job3/talend-job3.sh
How can I determine where the /scripts/talend.sh is started from, so I can CD to that path from within /scripts/talend.sh.
The Talend scripts are not run from within the directory itself, but from a cronjob, or a different users home directory.
EDIT:
The question was marked as duplicate, but Getting the source directory of a Bash script from within is not answering my question 100%.
Problem is:
- The basic script is being called from different scripts
- Those different scripts can be run from command line, with, and with or without a symbolic link.
- The $0, the $BASH_SOURCE and the pwd all do some things, but no solution mentioned covers all the difficulties.
Example:
/scripts/talend.sh
In this script I want to configure the $PATH and $HOME_PATH of Java, and CD to the place where the Talend job is placed. (It's a package, so that script MUST be run from that location).
Paths to the jobs are, for example:
/u/talend/talendjob1/sub../../talendjob1.sh
/u/talend/talendjob2/sub../../talendjob2.sh
/u/talend/talendjob3/sub../../talendjob3.sh
Multiple jobs are run from a TMS application. This application cannot run these scripts with the whol name (to long, name can only be 6 long), so in a different location I have symbolic links:
/u/tms/links/p00001 -> /u/talend/talendjob1/sub../../talendjob1.sh
/u/tms/links/p00002 -> /u/talend/talendjob1/sub../../talendjob2.sh
/u/tms/links/p00003 -> /u/talend/talendjob1/sub../../talendjob3.sh
/u/tms/links/p00004 -> /u/talend/talendjob1/sub../../talendjob4.sh
I think you get an overview of the complexity and why I want only one basic talend script, where I can leave all basic stuff. But I only can do that, if I know the source of the Talend script, because there I have to be to start that talend job.
These answers (beyond the first) are specific to Linux, but should be very robust there -- working with directory names containing spaces, literal newlines, wildcard characters, etc.
To change to your own source directory (a FAQ covered elsewhere):
cd "$(basename "$BASH_SOURCE")"
To change to your parent process's current directory:
cd "/proc/$PPID/cwd"
If you want to change to the directory passed as the first command-line argument to your parent process:
{ IFS= read -r -d '' _ && IFS= read -r -d '' argv1; } <"/proc/$PPID/cmdline"
cd "$argv1"
That said, personally, I'd just export the job directory to the environment variable in the parent process, and read that environment variable in the children. Much, much simpler, more portable, more accurate, and compliant with best process.
You can store pwd in a variable and then cd to it when you want to go back
This works for me:
In
/scripts/talend.sh
do
cd ${1%/*}
${1%/*} will strip off everything after the last / effectively providing a dirname for $1, which is the path to the script that calls this one.
and than call the script with the line:
/scripts/talend.sh $0.
Calling the script with $0 passes the name of the current script as an argument to the child which as shown above can be used to cd to the correct directory.
When you source /scripts/talend.sh the current directory is unchanged:
The scripts
# cat /scripts/talend.sh
echo "Talend: $(pwd)"
# cat /talend-job1_0.1/talend-job1_0.1/talend-job1/talend-job1.sh
echo Job1
. /scripts/talend.sh
Executing job1
# cd /talend-job1_0.1/talend-job1_0.1
# talend-job1/talend-job1.sh
Job1
Talend: /talend-job1_0.1/talend-job1_0.1
When you want to see the dir where the calling script is in, see get dir of script.
EDIT:
When you want to have the path of the callling script (talend-job1.sh) without having to cd to that dir first, you should get the dir of the script (see link above) and source talend.sh:
# cat /scripts/talend.sh
cd "$( dirname "${BASH_SOURCE[0]}" )"
echo "Talend: $(pwd)"
In talend.sh get the name of the calling script and then the directory:
parent_cmd=$(ps -o args= $PPID)
set -- $parent_cmd
parent_cmd=$(dirname $2)
Update: as pointed by Charles Duffy in the comments below this will cause havoc when used with paths containing white-space or glob patterns.
If procfs is available you could read the content of /proc/$PPID/cmdline or if portability is a concern do a better parsing of the args.
In /scripts/talend.sh:
cd "$(dirname "$0")"
Or:
cd "$(dirname "$BASH_SOURCE")"
Another one is:
cd "$(dirname "$_")"
#This must be the first line of your script after the shebang line
#Otherwise don't use it
Note: The most reliable of the above is $BASH_SOURCE
I'm searching for just one command — nothing with && or | — that creates a directory and then immediately changes your current directory to the newly-created directory. (This is a question someone got for his exams of "linux-usage", he made a new command that did that, but that didn't give him the points.) This is on a debian server if that matters.
I believe you are looking for this:
mkdir project1 && cd "$_"
define a bash function for that purpose in your $HOME/.bashrc e.g.
function mkdcd () {
mkdir "$1" && cd "$1"
}
then type mkdcd foodir in your interactive shell
So stricto sensu, what you want to achieve is impossible without a shell function containing some && (or at least a ; ) ... In other words, the purpose of the exercise was to make you understand why functions (or aliases) are useful in a shell....
PS it should be a function, not a script (if it was a script, the cd would affect only the [sub-] shell running the script, not the interactive parent shell); it is impossible to make a single command or executable (not a shell function) which would change the directory of the invoking interactive parent shell (because each process has its own current directory, and you can only change the current directory of your own process, not of the invoking shell process).
PPS. In Posix shells you should remove the functionkeyword, and have the first line be mkdcd() {
For oh-my-zsh users: take 'directory_name'
Reference: Official oh-my-zsh github wiki
Putting the following into your .bash_profile (or equivalent) will give you a mkcd command that'll do what you need:
# mkdir, cd into it
mkcd () {
mkdir -p "$*"
cd "$*"
}
This article explains it in more detail
I don't think this is possible but to all people wondering what is the easiest way to do that (that I know of) which doesn't require you to create your own script is:
mkdir /myNewDir/
cd !$
This way you don't need to write the name of the new directory twice.
!$ retrieves the last ($) argument of the last command (!).
(There are more useful shortcuts like that, like !!, !* or !startOfACommandInHistory. Search on the net for more information)
Sadly mkdir /myNewDir/ && cd !$ doesn't work: it retrieves the last of argument of the previous command, not the last one of the mkdir command.
Maybe I'm not fully understanding the question, but
>mkdir temp ; cd temp
makes the temp directory and then changes into that directory.
mkdir temp ; cd temp ; mv ../temp ../myname
You can alias like this:
alias mkcd 'mkdir temp ; cd temp ; mv ../temp ../'
You did not say if you want to name the directory yourself.
cd `mktemp -d`
Will create a temp directory and change into it.
Maybe you can use some shell script.
First line in shell script will create the directory and second line will change to created directory.